For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calculate the work of mechanically reversible, isothermal compression of 1 mol of methyl chloride from 1 bar to 55 bar at 100°C. Base calculations on the following forms of the virial equation: (a) Z = 1 + B __ V + C ___ V 2 (b) Z = 1 + B′P + C′ P 2 where B′ = B ___ RT and C′= C − B 2 ( RT ) 2 Why don’t both equations give exactly the same result?

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6  mo1^-2

Now according isothermal work of one mole methyl gas is

W=-

a=

b=

from virial equation  

And  

a=

b=

Now calculate V1 and V2 at given condition

Substitute given values = 1 x 10^5 , T = 373.15 and given values of coefficients we get  

Solve for V1 by iterative or alternative cubic equation solver we get

Similarly solve for state 2 at P2 = 50 bar we get  

Now  

a=241.33

b=30780

After performing integration we get work done on the system is  

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get  

         dV=RT(-1/p^2+0+C')dP

Hence work done on the system is  

a=

b=

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work  

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series  

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